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I'm learning solid angles from this page among others.

Half way down it shows a double integral:

$$\int_{\phi_1}^{\phi_2} d \phi\int_{\theta_1}^{\theta_2}\sin \theta\,d\theta = (\phi_2 - \phi_1)(\cos \theta_2 - \cos \theta_1)$$

I think the $\theta_1$ and $\theta_2$ on the RHS should be swapped. I'm very rusty on integration though. Can someone tell me what the right answer is please?

Hosein Rahnama
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PeteUK
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1 Answers1

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The result is wrong, and you are correct: $\theta_1$ and $\theta_2$ should be swapped.

We have $$\int^{\theta_2}_{\theta_1} \sin\theta\ d\theta = (-\cos\theta)\rvert^{\theta_2}_{\theta_1} = (-\cos\theta_2) - (-\cos\theta1) = \cos\theta_1 - \cos\theta_2.$$

Then $$\int_{\phi_1}^{\phi_2}d\phi\int^{\theta_2}_{\theta_1} \sin\theta\ d\theta = \int_{\phi_1}^{\phi_2}(\cos\theta_1 - \cos\theta_2)d\phi = (\phi_2 - \phi_1)(\cos\theta_1 - \cos\theta_2).$$

feralin
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